A) \[0.0195g\]
B) \[0.0013g\]
C) \[0.020g\]
D) \[0.0020g\]
Correct Answer: B
Solution :
From Faradays 1st law\[\omega =z.i.t\]. \[=\frac{E}{F}.q\] \[\left( \because \,\,\,Z=\frac{E}{F} \right)\] \[=\frac{31.75}{96500}\times 4\] \[=0.0013g\]
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